JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
If a semiconductor photodiode can detect a photon with a maximum wavelength of \(400\, nm ,\) then its band gap energy is Planck's constant \(h=6.63 \times 10^{-34}\, J . s\) Speed of ligh \(c =3 \times 10^{8}\, m / s\)
- A \(2.0\; eV\)
- B \(1.5\; eV\)
- C \(3.1\; eV\)
- D \(1.1 \;eV\)
Answer & Solution
Correct Answer
(C) \(3.1\; eV\)
Step-by-step Solution
Detailed explanation
\(\Delta E =\frac{h c }{\lambda e }=3.1\; eV\)
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